No. STP implies 0°C and 1 atm pressure, while standard conditions imply 25°C. We usually don't use STP outside of first-year chemistry, since most of our labs aren't that cold! Excellent question!
No. Looking at the structure of the molecule, it has one C and four H atoms. But each of those four C-H bonds has to be accounted for when calculating Delta H.
When calculating ΔH from enthalpies of formation it is. However, with bond enthalpies we calculate bonds broken minus bonds formed. This is the only time we take (left side) - (right side).
Good question. These bond enthalpy values refer to the amount of energy required to break a mole's worth of those bonds. Breaking bonds (on the reactant side) requires an input of energy, while forming bonds (on the product side) gives off energy. If we form bonds (releasing more energy) than we break (absorbing not as much energy), then the reaction will be exothermic. Thanks for watching!
haha i love how you throw in tricks to prepare us for the college boards tricks. thanks for the awesome content ur the best
Thanks! Over the years I've just about seen it all from CB, so I try to get students ready for the little 'twists' they like to throw in there.
thanks boss man
No problem!
Hi Mr. Krug! Is STP and standard conditions the same thing? If so, shouldn't it b 0 degrees celcius instead of 25 degrees?
No. STP implies 0°C and 1 atm pressure, while standard conditions imply 25°C. We usually don't use STP outside of first-year chemistry, since most of our labs aren't that cold! Excellent question!
Ok, Thank You!@@JeremyKrug
Wait so for the Ch4, wouldn't calculating 4 C-H bonds imply that there's 4 carbons and not just one?
No. Looking at the structure of the molecule, it has one C and four H atoms. But each of those four C-H bonds has to be accounted for when calculating Delta H.
isnt delta h products - reactants, not the other way around?
When calculating ΔH from enthalpies of formation it is. However, with bond enthalpies we calculate bonds broken minus bonds formed. This is the only time we take (left side) - (right side).
Why is it exothermic if the bonds of the products have more energy than the reactants
Good question. These bond enthalpy values refer to the amount of energy required to break a mole's worth of those bonds. Breaking bonds (on the reactant side) requires an input of energy, while forming bonds (on the product side) gives off energy. If we form bonds (releasing more energy) than we break (absorbing not as much energy), then the reaction will be exothermic. Thanks for watching!
YAY